The weight of a loaf of bakery bread at the grocery store follows a Normal distribution with a mean of µ = 12 ounces and a standard deviation of σ = 1 ounce. Suppose we pick four loaves at random from the bin andfind their total weight, W. Which of the following statements describes the random variable W?O The random variable W is binomial, with a mean of 48 ounces and a standard deviation of two ounces.O The random variable W is binomial, with a mean of 12 ounces and a standard deviation of two ounces.O The random variable W is Normal, with a mean of 12 ounces and a standard deviation of one ounce.O The random variable W is Normal, with a mean of 48 ounces and a standard deviation of two ounces.O The random variable W is Normal, with a mean of 48 ounces and a standard deviation of four ounces.

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Answered: The weight of a loaf of bakery bread at…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.25 probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 38.
The weight of a loaf of bakery bread at the grocery store follows a Normal distribution with a mean of μ = 12 ounces and a standard deviation of σ = 1 ounce. Suppose we pick four loaves at random from the bin and find their total weight, W. Which of the following statements describes the random variable W? O The random variable W is binomial, with a mean of 48 ounces and a standard deviation of two ounces. O The random variable W is binomial, with a mean of 12 ounces and a standard deviation of two ounces. O The random variable W is Normal, with a mean of 12 ounces and a standard deviation of one ounce. O The random variable W is Normal, with a mean of 48 ounces and a standard deviation of two ounces. O The random variable W is Normal, with a mean of 48 ounces and a standard deviation of four ounces.
Assume x is a normally distributed random variable with mean = 50 and variance = 3. Find the value of the random variable x0 when P(x ≤ x0) = .8413 A. 33 B. 49 C. 53 D. 16
6.2 The Standard Normal Distribution Table: Z-score and Probability A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 164.7 cm and a standard deviation of 2.5 cm. Find the probability that the length of a randomly selected steel rod is greater than 167.7 cm. Probability -
6
Let x be a random variable that represents the commute time of teachers in a large urban area. If x is normally distributed with a mean of 49 minutes and a standard deviation of 5.2 minutes, find the commute time which is at the top of the bottom 10%. A. 55.66 minutes B. 42.34 minutes C. 42.63 minutes D. 41.38 minutes
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The general philosophy of standardizing is that standardizing a random variable involves a. Subtracting the main in the batting by the standard deviation b. Dividing the standard deviation of the total of successes by the total of trials c. Adding and subtracting standard deviations in each direction d. Subtracting the area in the right tail from 1 since the Z chart gives only the area "to the left"
Suppose Y is normally distributed with mean = -3 and variance equal to 4. Find the probability that Y is less than or equal to -0.3. Hint, you must standardize the random variable. 0.9115 0.4602 0.5832 0.1
There are frogs and koi in a pond, and the number of frogs and the number of koi in the pond are independent. Let X represent the number of frogs in any given week, and let Y represent the number of koi in any given week. X has a mean of 28 with a standard deviation of 2.7, and Y has a mean of 15 with a standard deviation of 1.6. Which answer choice correctly calculates and interprets the standard deviation of the difference, D = X - Y? = 1.05; this pond can expect the difference of frogs and koi to vary by approximately 1.05 from the mean. = 1.1; this pond can expect the difference of frogs and koi to vary by approximately 1.1 from the mean. = 3.1; this pond can expect the difference of frogs and koi to vary by approximately 3.1 from the mean. = 13; this pond can expect the difference of frogs and koi to vary by approximately 1.05 from the mean.
Assume the random variable X is normally distributed with mean u = 50 and standard deviation o = 7. Find the 77th percentile. The 77th percentile is: (Round to two decimal places as needed.)
If one normal distribution has a standard deviation of 5, and another has a standard deviation of 25, both still contain 68% of their values within one standard deviation. O True O False MCD +2.09% o search Del F10 F11 F12 Prt Sc Insert F9 F8 + F6 F7 F3 F4 F5 Baskspace

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